I've got an idea about the first 1, but not 100% on it. I think that the derivative of a function is 1/the derivative of the inverse function, so just calculate the df/dx for x = 3, then the answer is the reciprocal of the number you get.

For the second question, ever thought about writing it as [itex] \int 1 dx - \int \frac{4}{x+4} dx [/itex] ?

I assume that by (f-1) you mean the derivative of 1/f(x) (because if it were the inverse, the exercise would become quite messy and ugly). You can do this in two ways (three, actually):
a) Write [tex]1/\sqrt{\cdots} = (\cdots)^{-1/2}[/tex] and use the familiar rule.
b) Use the quotient rule
c) Use a rule for [tex]\frac{d(1/f(x))}{dx}[/tex] if you know it (otherwise you can derive it using b)

I'm not entirely sure about the first part, but I think that if you find the derivative for f, then put 3 into the resulting equation, and then simply do (your answer)^-1 you will get the derivative of the function f^-1 at x = 3.

The second part, all I did was break the integral up. Is the following not true?
[tex] 1 - \frac{4}{x+4} = \frac{x}{x+4} [/tex]
If it is, then the integral of one (with respect to x), must be the integral of the other (w.r.t x). I just transformed the integral by inspection, there isn't anything special going on there.

I'm not entirely sure about the first part, but I think that if you find the derivative for f, then put 3 into the resulting equation, and then simply do (your answer)^-1 you will get the derivative of the function f^-1 at x = 3.

That is not true.
As a counterexample, consider
[tex]f(x) = x; g(x) = f(x)^{-1} = \frac{1}{x}[/tex].
Clearly, [itex]f'(x) = 1[/itex], so by your argument we should have [itex]g'(x) = 1 / f'(x) = 1 / 1 = 1[/itex] for all x.
But actually,
[tex]g'(x) = - \frac{1}{x^2}[/tex]
which is in general not equal to 1 (in fact, if x is real, it will never hold).

You really need the quotient rule or "power rule" (I don't know what it's called - if it has a name - I mean the rule dx^n/dx = n x^(n-1)) to find the derivative of 1/f(x).

That is not true.
As a counterexample, consider
[tex]f(x) = x; g(x) = f(x)^{-1} = \frac{1}{x}[/tex].
Clearly, [itex]f'(x) = 1[/itex], so by your argument we should have [itex]g'(x) = 1 / f'(x) = 1 / 1 = 1[/itex] for all x.
But actually,
[tex]g'(x) = - \frac{1}{x^2}[/tex]
which is in general not equal to 1 (in fact, if x is real, it will never hold).

You really need the quotient rule or "power rule" (I don't know what it's called - if it has a name - I mean the rule dx^n/dx = n x^(n-1)) to find the derivative of 1/f(x).